[英]Why Is jQuery Not Submitting This Form? (Or What Have I Done Wrong? :-p)
[英]PHP Ajax jQuery Post - What have I done wrong?
這是我的 jQuery...
$.ajax({
type: 'POST',
url: 'http://adamscarter.co.uk/daily/facebook/savefbprofile.php',
data: {'fb_id': fb_id, 'fbusername': fbusername, 'location_id': location_id, 'gender': gender},
beforeSend: function() {
console.log('Before send: ' + fb_id, fbusername, location_id, gender);
},
success: function(data) {
console.log('saveFbProfile post ajax success');
console.log(data);
}
});
...這是我的 PHP ...
<?php
session_start();
include('dbcon.php');
$gender = substr($_POST['gender'], 0, 1);
$fb_id = $_POST['fb_id'];
$location_id = $_POST['location_id'];
$fbusername = $_POST['fbusername'];
//Set session vars
$_SESSION['gender'] = $gender;
$_SESSION['fb_id'] = $fb_id;
$_SESSION['location_id'] = $location_id;
$_SESSION['fbusername'] = $fbusername;
if (isset($_SESSION['userid'])) {
mysql_select_db('users', $GLOBALS['conInsert']);
$sql = "UPDATE user_fb_details SET gender = '". $gender. "', fb_id = '". $fb_id. "', location_id = '". $location_id. "', username = '". $fbusername. "' WHERE user_id = '". $_SESSION['userid']. "'";
mysql_query($sql, $GLOBALS['conInsert']);
}
?>
我的代碼有錯誤嗎? 當我記錄變量時,它們都有正確的值,但我只是得到一個“parseerror”。
聽起來 jQuery 可能正試圖將 POST 的結果解析為 JSON 響應並失敗,因為它不是 JSON。 您可能希望將dataType設置為'text'
,或從 PHP 返回不同的Content-Type
。
你肯定想用一些mysql_real_escape_string
或參數化查詢來修復危險的 SQL 注入漏洞。
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